From 03c493dd0efd8de6b5bd5bf27af63654146e4cfe Mon Sep 17 00:00:00 2001
From: Phillip Johnston <phillip@embeddedartistry.com>
Date: Wed, 27 Oct 2021 10:18:09 -0700
Subject: [PATCH] Update shared pointer example to use make_shared

Fixes #29
---
 examples/cpp/shared_ptr.cpp | 19 ++++++++++++++++++-
 1 file changed, 18 insertions(+), 1 deletion(-)

diff --git a/examples/cpp/shared_ptr.cpp b/examples/cpp/shared_ptr.cpp
index 9b34ad2..18d5f69 100644
--- a/examples/cpp/shared_ptr.cpp
+++ b/examples/cpp/shared_ptr.cpp
@@ -29,7 +29,24 @@ class SharedThing : public std::enable_shared_from_this<SharedThing>
 	template<typename... T>
 	static std::shared_ptr<SharedThing> create(T&&... t)
 	{
-		return std::shared_ptr<SharedThing>(new SharedThing(std::forward<T>(t)...));
+		// C++ core guidelines say we shouldn't do this:
+		// return std::shared_ptr<SharedThing>(new SharedThing(std::forward<T>(t)...));
+
+		// But this doesn't work because our constructor is private, and we want that
+		// for use with enable_shared_from_this
+		// return std::make_shared<SharedThing>(std::forward<T>(t)...);
+
+		// So we'll make a "fake" derived class that constructs our shared thing
+		// using a public constructor...
+		struct EnableMakeShared : public SharedThing
+		{
+			EnableMakeShared(T&&... arg) : SharedThing(std::forward<T>(arg)...)
+			{
+			}
+		};
+
+		// and use that type with make_shared
+		return std::make_shared<EnableMakeShared>(std::forward<T>(t)...);
 	}
 
   private: